Animated Solution for Mathematics - Trigonometry: The number of elements in the set $S=\left\{x\in \mathbb{R}:2\cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\right\}$ is:

Imagine you are standing at the crossroads of two mathematical worlds. On one side, you have the rhythmic, oscillating waves of trigonometry. On the other, the explosive, rapid growth of exponential functions.

When you see an equation like , your first instinct might be to panic. How do you equate a wave to an exponential curve?

The secret is not to fight the equation, but to observe it. This is a classic JEE Advanced trap designed to test your ability to look beyond the algebra and into the soul of the functions.

Let us isolate the Right Hand Side (RHS): $4^x+4^{-x}$. This is a beautiful structure. Notice that $4^x$ is always positive for any real $x$.

Therefore, we have a sum of two positive numbers that are reciprocals of each other. Whenever you see this, the Arithmetic Mean - Geometric Mean (AM-GM) inequality should immediately flash in your mind.

The AM-GM inequality states that for any two positive numbers $a$ and $b$:

Let $a=4^x$ and $b=4^{-x}$. Substituting these into the inequality, we get:

Since $4^x\cdot 4^{-x}=4^{x-x}=4^0=1$, the inequality simplifies to:

The RHS is strictly bounded below by 2. It can never be less than 2.

Now, let us turn our attention to the Left Hand Side (LHS): $2\cos \left(\frac{x^2+x}{6}\right)$. The expression inside the cosine might look like a terrifying quadratic, but remember the fundamental property of the cosine function: for any real angle $\theta$, $-1\le \cos (\theta )\le 1$.

It does not matter if the input is a simple $x$ or a complex quadratic; the output is always trapped between -1 and 1. When we multiply this by 2, the entire range is scaled:

The maximum value the LHS can ever achieve is exactly 2.

We have reached the climax of our journey. We know that the RHS is always $\ge 2$, and the LHS is always $\le 2$. The equation demands that they be equal.

The only way for a value that is at least 2 to equal a value that is at most 2 is if both sides are exactly 2 at the same time. This is the 'Aha!' moment.

For the RHS, equality in AM-GM holds if and only if the terms are equal:

Now, we must verify this in the LHS. Substituting $x=0$ into $2\cos \left(\frac{x^2+x}{6}\right)$, we get:

It works! Both sides meet at the value 2 when $x=0$.

The Elegance of the Solution

We have found that $x=0$ is the only point of intersection. The set $S$ contains exactly one element.

This problem is a masterclass in why we study functions. By understanding the boundaries of our tools, we can solve problems that seem impossible at first glance.

You did not need a calculator or complex calculus; you just needed to understand the limits of the functions. Keep this mindset, and you will conquer any problem the JEE throws at you.